Problem 47 - Thomas Calculus

Question

Quadratic Approximations The Taylor polynomial of order 2 generated by a twice-differentiable function \(f(x)\) at \(x=a\) is called the quadratic approximation of \(f\) at \(x=a .\) Find the (a) linearization (Taylor polynomial of order 1 ) and (b) quadratic approximation of \(f\) at \(x=0\). \(f(x)=1 / \sqrt{1-x^{2}}\)

Step-by-Step Solution

Answer

Linearization: 1. Quadratic approximation: \(1 + \frac{x^2}{2}\).

Step 1: Find f(0), f'(0), f''(0)

\(f(x) = (1-x^2)^{-1/2}\)
\(f(0) = 1\)
\(f'(x) = x(1-x^2)^{-3/2}\), \(f'(0) = 0\)
\(f''(x) = (1-x^2)^{-3/2} + 3x^2(1-x^2)^{-5/2}\), \(f''(0) = 1\)

Step 2: Linearization (Order 1)

\(P_1(x) = f(0) + f'(0)x = 1\)

Step 3: Quadratic Approximation (Order 2)

\(P_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 = 1 + \frac{1}{2}x^2\)

Key Concepts

Quadratic Approximation

The quadratic approximation is a simple and powerful way to approximate a function around a specific point. It uses a Taylor polynomial of degree two. When you have a function, such as \( f(x) \), you can expand it using the Taylor polynomial to express it in the form:

• \( f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 \)

This is especially useful for functions that are not easy to compute directly close to a point like \( x = a \). By using the quadratic approximation, you can simplify these computations. The result gives you an estimate of how the function behaves near that point. This is particularly effective when dealing with complex functions in calculus, like \( f(x) = \frac{1}{\sqrt{1-x^2}} \). Knowing how to find the first and second derivatives, as shown in the solution steps, is key to applying this approximation.

Linearization

Linearization, or the Taylor polynomial of order 1, is the simplest form of approximating a function near a point. It gives a straight line that tangentially touches the function at a particular point \( x = a \). Mathematically, it’s expressed as:

• \( f(x) \approx f(a) + f'(a) \cdot (x-a) \)

This approximation is valuable because it provides a linear function that you can calculate easily, simplifying complex function behavior in that immediate vicinity. For \( f(x) = \frac{1}{\sqrt{1-x^2}} \), by evaluating the linearization at \( x = 0 \), you get the slope (i.e., derivative) at that point, which helps in understanding the function's behavior very close to zero with minimal calculation.

Derivatives

Derivatives are at the heart of finding both linear and quadratic approximations. A derivative reflects how a function changes as its input changes, providing insights into the function’s behavior. The first derivative \( f'(x) \) tells us about the slope or the rate of change at any given point. To find it for a function like \( f(x) = \frac{1}{\sqrt{1-x^2}} \), you apply differentiation rules, like the chain rule. After that, the second derivative \( f''(x) \) offers even deeper insight into the curvature and concavity of the function. For instance, in our exercise, determining \( f''(x) \) required application of both the chain rule and quotient rule for proper calculation, leading to a more comprehensive understanding of how the function bends around the tested point \( x = 0 \).

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate compositions of functions. When a function is composed of another function, like a nested set, the chain rule helps us find the derivative of the outer function and multiply it by the derivative of the inner function.

• If \( y = g(h(x)) \), then \( \frac{dy}{dx} = g'(h(x)) \cdot h'(x) \).

Applied to \( f(x) = \frac{1}{\sqrt{1-x^2}} \), the chain rule becomes vital because it allows you to treat everything inside the square root as a single function. You then differentiate it as such, which simplifies finding \( f'(x) \). By breaking the derivation process into manageable steps, you efficiently arrive at the result without losing sight of the functions' relationships.

Quotient Rule

The quotient rule is an essential differentiation technique when you have a ratio of two functions. It helps determine the derivative of a fraction’s numerator divided by its denominator function. The rule states:

• \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \)

For functions like \( f(x) = \frac{x}{(1-x^2)^{\frac{3}{2}}} \), which involve division, the quotient rule is necessary. Applying it requires separately differentiating both the top function (u) and bottom function (v), and then combining these using the quotient formula. In the exercise, the quotient rule helped with finding \( f''(x) \) by distinguishing between the variable and its divisor, thus leading to a precise calculation of the second derivative essential for accurate quadratic approximation.